Reducing $\cos(2\pi/3)$, $\tan(7\pi/4)$, and $\sin(7\pi/6)$

Question

It's been a few years since doing any type of trigonometry questions and I've seemed to forgotten everything about it. Below is a question with the solution. You're not supposed to use a calculator.

$$\begin{align} &\cos\frac{2\pi}{3}+\tan\frac{7\pi}{4}-\sin\frac{7\pi}{6} \\[4pt] &=-\cos\frac{\pi}{3}-\tan\frac{\pi}{4}-\left(-\sin\frac{\pi}{6}\right) \\[4pt] &=-\frac12-1+\frac12 \\[4pt] &=-1 \end{align}$$

Can somebody explain the following to me?

• How $\cos(2\pi/3)$ becomes $-\cos(\pi/3)$
• How $\tan(7\pi/4)$ becomes $-\tan(\pi/4)$
• How $-\sin(7\pi/6)$ becomes $-(-\sin(\pi/6))$

Thanks

Answer 1

Notice that $\cos{(2 \pi / 3)}$ is negative since it is located in the 2nd quadrant. Since $\cos{( \pi - 2 \pi /3 )}$ has the same value as $\cos{(2 \pi / 3)}$ in absolute values (i.e. $\cos{( \pi - 2 \pi /3 )}$ is positive) you can rewrite it as: $$ \cos{(2 \pi / 3)} = - \cos{( \pi - 2 \pi /3 )} =- \cos{( \pi /3)} $$ A similar reasoning applies to the other examples.

Answer 2

Using the following identities:

$$\cos(\pi - \alpha) = -\cos(\alpha)$$ $$\tan(\alpha + 2\pi) = \tan(\alpha)$$ $$\tan(-\alpha) = -\tan(\alpha)$$ $$\sin(\alpha + \pi) = -\sin(\alpha)$$

We get:

$$\cos\dfrac{2\pi}{3} = \cos\left(\pi - \dfrac{\pi}{3}\right) = -\cos\dfrac{\pi}{3}$$

$$\tan\dfrac{7\pi}{4} = \tan\left(-\dfrac{\pi}{4} + 2\pi\right) = \tan\left(-\dfrac{\pi}{4}\right) = -\tan\dfrac{\pi}{4}$$

$$-\sin\dfrac{7\pi}{6} = -\sin\left(\dfrac{\pi}{6} + \pi\right) = -\left(-\sin\dfrac{\pi}{6}\right)$$